SURF matching

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km g
km g 2011-8-15
SURF descriptor will extract one vector with length 64. If i want to match between two regions, how am i going to match this two descriptor?i search a lot of information and mostly they use euclidean distance to get the value. if i use euclidean distance, how can i going to do that? by minus each 64 value with other 64 values in the other region? or each one value compare with other 64 value?if 64 * 64 =4096, then the process will be very time consuming! anyone pls answer me?!
  2 个评论
Jan
Jan 2011-8-15
Please post more details, most of all the code you have created already.
km g
km g 2011-8-16
length=20*s/2;
g=fspecial('gaussian',[(2*length)+1 (2*length)+1],3.3*s);
theta=-direction*pi/180;
[diy,dix]=ndgrid(-length-1:length+1,-length-1:length+1);
test(1:200,1:200)=0;
for m=1:size(dix,1)
for n=1:size(dix,2)
test(r+diy(m,n),c+dix(m,n))=z1(r+diy(m,n),c+dix(m,n));
end
end
for m=1:size(diy,1)
for n=1:size(diy,2)
newx(m,n)=(dix(m,n)*cos(theta))-(diy(m,n)*sin(theta));
newy(m,n)=(dix(m,n)*sin(theta))+(diy(m,n)*cos(theta));
end
end
cox=round(c+newx);
coy=round(r+newy);
cox=cox(:);
coy=coy(:);
test2(1:200,1:200)=0;
for cc=1:size(cox,1)
newmapping(cc)= z1(coy(cc),cox(cc));
test2(coy(cc),cox(cc))=z1(coy(cc),cox(cc));
end
newmapping=reshape(newmapping,23,23);
figure,imshow(test,[]);
figure,imshow(test2,[]);
figure,imshow(newmapping,[]);
% g=g(:);
sizec=2*s;
sizer=2*s;
%
% nhaarx(1:200,1:200)=NaN;
% nhaary(1:200,1:200)=NaN;
for cc=2:22
for rr=2:22
nhaarx(rr-1,cc-1)=(boxintegral(rr-sizer/2,cc,newmapping,sizec/2,sizer)-boxintegral(rr-sizer/2,cc-sizec/2,newmapping,sizec/2,sizer));
nhaary(rr-1,cc-1)=(boxintegral(rr,cc-sizec/2,newmapping,sizec,sizer/2)-boxintegral(rr-sizer/2,cc-sizec/2,newmapping,sizec,sizer/2));
end
end
%
% figure,imshow(nhaarx,[]);
% figure,imshow(nhaary,[]);
for m=1:size(nhaarx,1)
for n=1:size(nhaary,2)
if dix(m,n)==0 || diy(m,n)==0
nhaarx(m,n)=NaN;
nhaary(m,n)=NaN;
end
end
end
q=1;
for m=1:size(nhaarx,1)
for n=1:size(nhaarx,2)
if isnan(nhaarx(m,n))==0 && isnan(nhaary(m,n))==0
newregionxx(q,1)=nhaarx(m,n);
newregionyy(q,1)=nhaary(m,n);
q=q+1;
end
end
end
si=sqrt(size(newregionxx,1)*size(newregionxx,2));
newregionx=reshape(newregionxx(:,1),si,si);
newregiony=reshape(newregionyy(:,1),si,si);
figure,imshow(newregionx,[]);
figure,imshow(newregiony,[]);
p=1;
gap=20*s/4;
for m=1:gap:size(newregionx,1)
for n=1:gap:size(newregiony,1)
[sumx,sumax,sumy,sumay]=getvector(m,n,gap,newregionx,newregiony);
sx(p)=sumx;
sy(p)=sumy;
sax(p)=sumax;
say(p)=sumay;
p=p+1;
end
end
descriptorv(1:16)=sx(:);
descriptorv(17:32)=sy(:);
descriptorv(33:48)=sax(:);
descriptorv(49:64)=say(:);
descriptorvv(1:16)=sy(:);
descriptorvv(17:32)=sx(:);
descriptorvv(33:48)=say(:);
descriptorvv(49:64)=sax(:);

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回答（1 个）

David Young
David Young 2011-8-15
The Euclidean distance between vectors A and B is computed simply using
norm(A-B)
This applies to SURF descriptor vectors too.
  5 个评论
显示 3更早的评论
David Young
David Young 2011-8-17
You can either rotate the image patch, or you can rotate the kernels, but not both.
km g
km g 2011-8-18
if i rotate the image patch, then do i need to compute the integral image again for the rotated image patch?or i directly use the integral image of that part by using the one that compute earlier?

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